To solve for the forces/speed of elevator, look no further than reducing everything to free body diagram(s). Remember in a free body diagram we want everything to equal zero in static calculations (which we have simplified this to). We are neglecting the weight of the elevator parts in these diagrams. Also, we are assuming all rope forces are perfectly vertical (which they should be if you designed the elevator right).

**Continuous:**

The free body diagram of continuous elevators are relatively simple; the cable in this effectively just keeps toggling the weight on the end of the elevator up and down again. I broke this down into three notes, N1, N2, N3.

Node 1 is the top pulley of the second stage, Node 2 is the bottom pulley of the second stage, and Node 3 is the top pulley of the first (fixed stage).

Node 1 sees the weight on the end of the elevator, expressed as W. Both sides of a cable around a pulley see identical force. The position of the elevator is static, so the cable itself must exert equal and opposite forces, which means they both point down (since the cable is curved around a pulley with 180 degrees of wrap). To keep the node static, the pulley exerts a +2W to counter-act the -2W force exerted by both sides of the cable.

Node 2 is the equal and opposite version of Node 1, and Node 3 is the equal and opposite of Node 2 (and identical to Node 1). The weight of the elevator load travels entirely through the rope, and at any given point in the cable the magnitude of the force is always equal to exactly W, no more, no less. Note in this version the +2W and -2W forces are entirely internal (they only apply to the fixed members inside the elevator).

Thus, the cable at the motor sees a force equal to that weight of the object, and the object travels upward at a 1:1 rate compared to how much the cable is pulley.

**Cascade:**

Cascade is a bit different. Because there are multiple cables, we are not guaranteed the same force in the cable at any given point. So we need to be more careful with this version on setting up the FBD correctly. Here is the drawing:

Node 1 starts the same as continuous, with two separate -W forces and one +2W force inside the second stage beam. But Node 2 is where things change.

Node 2 does not see any direct rope forces from Node 1; the rope from Node 1 is tied to a fixed point on the first stage. Node 2 sees the internal force of the second stage fixed member, then uses the second rope force to equalize this. The -2W force is the equal and opposite force to the +2W force seen in Node 1 transmitted through the rigid elevator member. To keep this node static, we need the cable to Node 3 to transmit a +2W force.

Node 3 is contains the -2W cable force from Node 2 which also means a 2W force going to the motor, which means we need a total of +4W vertical force (1W of this force will be from the cable from Node 1) at the pulley to counter act it.

Thus, the cable at the motor sees 2x the weight of the object. But for every centimeter that the second stage is directly pulled upwards by the motor, the third stage carriage moves up at 2x this rate.

All said and done, these both use the same motor mechanical power to move the final (third) stage. Using continuous as a baseline, cascade requires 2x the torque at the input, but only requires half the input speed to make the output stage move at the same rate*.

- These only hold true if there are two moving stages.